Two events are independent if the following are true:
- P(A|B) = P(A)P(A|B) = P(A)
- P(B|A) = P(B)P(B|A) = P(B)
- P(A AND B) = P(A) ⋅ P(B)P(A AND B) = P(A) ⋅ P(B)
If AA and BB are independent, then the chance of AA occurring does not affect the chance of
BB occurring and vice versa. For example, two roles of a fair die are independent events.
The outcome of the first roll does not change the probability for the outcome of the second
roll. To show two events are independent, you must show only one of the above conditions.
AA and BB are mutually exclusive events if they cannot occur at the same time. This means
that AA and BB do not share any outcomes and P(A AND B) = 0P(A AND B)= 0.
For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}C = {7, 9}. A AND B = {4, 5}A AND B ={4, 5}. P(A AND B) = P(A AND B) =
2
10
2
10
and is not equal to zero. Therefore, AA and BB are not mutually exclusive. AA and CC do not have any numbers in common so
P(A AND C) = 0P(A AND C) = 0. Therefore, AA and CC are mutually exclusive.
Independent and mutually exclusive do not mean the same thing.
You must show that any two events are independent or mutually exclusive. You
cannot assume either of these conditions.
If it is not known whether AA and BB are independent or dependent, assume they
are dependent until you can show otherwise.
The following examples illustrate these definitions and terms.
Flip two fair coins. (This is an experiment.)
The sample space is {HH, HT, TH, TT}{HH, HT, TH, TT} where TT = tails and HH = heads. The outcomes
are HHHH, HTHT, THTH, and TTTT. The outcomes HTHT and THTH are different. The HTHT means that
the first coin showed heads and the second coin showed tails. The THTH means that the first coin showed tails and the second coin showed heads.
- Let AA = the event of getting at most one tail. (At most one tail means 0 or 1 tail.)
Then AA can be written as {HH, HT, TH}{HH, HT, TH}. The outcome HHHH shows 0 tails. HTHT and
THTH each show 1 tail.
- Let BB = the event of getting all tails. BB can be written as {TT}{TT}. BB is the complement of
AA. So, B = A'B = A'. Also, P(A) + P(B) = P(A) + P(A') = 1P(A) + P(B) = P(A) + P(A') = 1.
- The probabilities for AA and for BB are P(A) =
3
4
P(A) =
3
4
and P(B) =
1
4
P(B)=
1
4
.
- Let CC = the event of getting all heads. C = {HH}C = {HH}. Since B = {TT}B = {TT}, P(B AND C) = 0P(B AND C) = 0. BB
and CC are mutually exclusive. (BB and CC have no members in common because you
cannot have all tails and all heads at the same time.)
- Let DD = event of getting more than one tail. D = {TT}D ={TT}. P(D) =
1
4
P(D)=
1
4
.
- Let EE = event of getting a head on the first roll. (This implies you can get either a head
or tail on the second roll.) E = {HT, HH}E={HT, HH}. P(E) =
2
4
P(E) =
2
4
.
- Find the probability of getting at least one (1 or 2) tail in two flips. Let FF = event of
getting at least one tail in two flips. F= {HT, TH, TT}F={HT, TH, TT}. P(F) =
3
4
P(F) =
3
4
Roll one fair 6-sided die. The sample space is {1, 2, 3, 4, 5, 6}{1, 2, 3, 4, 5, 6}.
Let event AA = a face is odd. Then A = {1, 3, 5}A = {1, 3, 5}. Let event BB = a face is even.
Then B = {2, 4, 6}B = {2, 4, 6}.
- Find the complement of AA, A'A'. The complement of AA, A'A', is BB because AA and
BB together make up the sample space. P(A) + P(B) = P(A) + P(A') = 1P(A) + P(B) = P(A) + P(A') = 1.
Also, P(A) =
3
6
P(A)=
3
6
and P(B) =
3
6
P(B)=
3
6
- Let event CC = odd faces larger than 2. Then C = {3, 5}C={3,5}. Let event DD = all even
faces smaller than 5. Then D = {2, 4}D ={2,4}. P(C and D) = 0P(C and D) =0 because you cannot have an
odd and even face at the same time. Therefore, CC and DD are mutually exclusive
events.
- Let event EE = all faces less than 5. E = {1, 2, 3, 4}E ={1,2,3,4}.
Are CC and EE mutually
exclusive events? (Answer yes or no.) Why or why not?
No. CC = {3, 5}{3, 5} and EE = {1, 2, 3, 4}{1, 2, 3, 4}. P(C AND E)=16P(C AND E)=16. To be mutually exclusive, P(C AND E)P(C AND E) must be 0.
- Find P(C|A)Find P(C|A). This is a conditional. Recall that the event CC is
{3, 5}{3, 5}
and event AA is
{1, 3, 5}{1, 3, 5}. To find P(C|A)P(C|A), find the probability of CC using the sample space AA. You
have reduced the sample space from the original sample space
{1, 2, 3, 4, 5, 6}{1, 2, 3, 4, 5, 6}
to {1, 3, 5}{1, 3, 5}. So, P(C|A) =
2
3
P(C|A) =
2
3
Let event GG = taking a math class. Let event HH = taking a science class. Then, G AND HG AND H = taking a math class and a science class. Suppose P(G) = 0.6P(G) =0.6, P(H)= 0.5P(H)= 0.5, and P(G AND H) = 0.3P(G AND H) = 0.3. Are GG and HH independent?
If GG and HH are independent, then you must show ONE of the following:
- P(G|H) = P(G)P(G|H) = P(G)
- P(H|G) = P(H)P(H|G) = P(H)
- P(G AND H)=P(G AND H)= P(G)⋅P(H)P(G)⋅P(H)
The choice you make depends on the information you have. You could choose any of
the methods here because you have the necessary information.
Show that
P(G|H) = P(G)P(G|H) = P(G).
P(G|H)
=
P(G AND H)
P(H)
=
0.3
0.5
=
0.6
=
P(G)
P(G|H) =
P(G AND H)
P(H)
=
0.3
0.5
= 0.6 = P(G)
Show P(G AND H)=P(G AND H)= P(G)⋅P(H)P(G)⋅P(H).
P(G)
⋅
P(H)
=
0.6
⋅
0.5
=
0.3
=
P(G AND H)
P(G)⋅P(H) = 0.6⋅0.5 = 0.3 = P(G AND H)
Since GG and HH are independent, then, knowing that a person is taking a science class does
not change the chance that he/she is taking math. If the two events had not been
independent (that is, they are dependent) then knowing that a person is taking a science
class would change the chance he/she is taking math. For practice, show that P(H|G) = P(H)P(H|G) = P(H) to show that GG and HH are independent events.
In a box there are 3 red cards and 5 blue cards. The red cards are
marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3,
4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and
draw one card.
Let RR = red card is drawn, BB = blue card is drawn, EE = even-numbered card is drawn.
The sample space
S
=
R1, R2, R3, B1, B2, B3, B4, B5
S = R1, R2, R3, B1, B2, B3, B4, B5. SS has 8 outcomes.
-
P(R)
=
3
8
P(R)=
3
8
.
P(B)
=
5
8
P(B)=
5
8
.
P(R AND B)
=
0
P(R AND B)=0. (You cannot draw one card that is both
red and blue.)
- P(E) =
3
8
P(E)=
3
8
. (There are 3 even-numbered cards, R2R2, B2B2, and B4B4.)
- P(E|B) =
2
5
P(E|B)=
2
5
. (There are 5 blue cards: B1B1, B2B2, B3B3, B4B4, and B5B5. Out of the blue cards,
there are 2 even cards: B2B2 and B4B4.)
- P(B|E) =
2
3
P(B|E)=
2
3
. (There are 3 even-numbered cards: R2R2, B2B2, and B4B4. Out of the
even-numbered cards, 2 are blue: B2B2 and B4B4.)
- The events RR and BB are mutually exclusive because P(R AND B) = 0P(R AND B) = 0.
- Let GG = card with a number greater than 3. G = {B4, B5}G ={B4,B5}. P(G) =
2
8
P(G)=
2
8
.
Let HH = blue card numbered between 1 and 4, inclusive. H = {B1,B2,B3, B4}H ={B1,B2,B3,B4}.
P(G|H) =
1
4
P(G|H)=
1
4
. (The only card in H that has a number greater than 3 is B4B4.)
Since
2
8
=
1
4
2
8
=
1
4
, P(G) = P(G|H)P(G) =P(G|H) which means that GG and HH are independent.
- Independent Events:
The occurrence of one event has no effect on the probability of the occurrence of any other event. Events A and B are independent if one of the following is true: (1). P
(
A
2
B
)
=
P
(
A
)
;P(
A
2
B)=P(A); (2) P
(
B
2
A
)
=
P
(
B
)
;P(
B
2
A)=P(B); (3) P
(
A
and
B
)
=
P
(
A
)
P
(
B
)P(AandB)=P(A)P(B).
- Mutually Exclusive:
An observation cannot fall into more than one class (category). Being in one category prevents being in a mutually exclusive category.
"This is the course textbook for Biology 502 at CSU Dominguez Hills"